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Correct way to find the area of a concave quadrilateral in co-ordinate geometry using triangles.
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A lot of textbooks where I live teach students to calculate the area of a quadrilateral using its coordinates by considering it to be made up of two triangles.
We calculate the area of each triangle using the following formula:
|(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|/2
Their individual areas are then added to calculate the final value. Won't this be a problem when calculating the area of concave quadrilaterals?
Is there any way to make sure that the area calculated using this method is indeed the correct area?
geometry analytic-geometry coordinate-systems
asked Jul 23 at 16:52
iKnowNothing
84
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up vote
1
down vote
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A lot of textbooks where I live teach students to calculate the area of a quadrilateral using its coordinates by considering it to be made up of two triangles.
We calculate the area of each triangle using the following formula:
|(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|/2
Their individual areas are then added to calculate the final value. Won't this be a problem when calculating the area of concave quadrilaterals?
Is there any way to make sure that the area calculated using this method is indeed the correct area?
geometry analytic-geometry coordinate-systems
asked Jul 23 at 16:52
iKnowNothing
84
the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A lot of textbooks where I live teach students to calculate the area of a quadrilateral using its coordinates by considering it to be made up of two triangles.
We calculate the area of each triangle using the following formula:
|(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|/2
Their individual areas are then added to calculate the final value. Won't this be a problem when calculating the area of concave quadrilaterals?
Is there any way to make sure that the area calculated using this method is indeed the correct area?
geometry analytic-geometry coordinate-systems
asked Jul 23 at 16:52
iKnowNothing
84
A lot of textbooks where I live teach students to calculate the area of a quadrilateral using its coordinates by considering it to be made up of two triangles.
We calculate the area of each triangle using the following formula:
|(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|/2
Their individual areas are then added to calculate the final value. Won't this be a problem when calculating the area of concave quadrilaterals?
Is there any way to make sure that the area calculated using this method is indeed the correct area?
geometry analytic-geometry coordinate-systems
asked Jul 23 at 16:52
iKnowNothing
84
asked Jul 23 at 16:52
iKnowNothing
84
asked Jul 23 at 16:52
iKnowNothing
84
asked Jul 23 at 16:52
iKnowNothing
84
84
the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32
add a comment |Â
the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32
the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32
add a comment |Â
2 Answers
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oldest
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up vote
0
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accepted
If you use the formula given as sum of products, then this is a signed area. That is, depending on the order of the three points, you may get a negative value. The formula generalizes to any polygon. This is sometimes known as the shoelace formula. As the Wikipedia article states "The area formula is valid for any non-self-intersecting (simple) polygon, which can be convex or concave". Using the general formula, the absolute value of the signed area is the area you wanted.
Let the point coordinates be
$ P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4). $
The area of triangle $ P_1P_2P_4 $ is
$ A := x_1(y_2-y_4) + x_2(y_4-y_1) + x_4(y_1-y_2). $ The area of triangle
$ P_4P_2P_3 $ is
$ B := x_4(y_2-y_3) + x_2(y_3-y_4) + x_3(y_4-y_2). $ Add the two signed areas to get $ A + B = (x_1y_2-x_2y_1) + (x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) + (x_4y_1-x_1y_4) $ from shoelace.
answered Jul 23 at 17:54
Somos
11.4k1833
add a comment |Â
up vote
0
down vote
You can surely use that method but it would be quite "bashy". Especially if you do it by hand. A better tool for this case (or any polygon) would be Gauss's Area Theorem, AKA shoelace theorem. I think this page provides a better explanation if shoelace theorem than Wikipedia.
https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem
answered Jul 23 at 18:01
Shrey Joshi
1389
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If you use the formula given as sum of products, then this is a signed area. That is, depending on the order of the three points, you may get a negative value. The formula generalizes to any polygon. This is sometimes known as the shoelace formula. As the Wikipedia article states "The area formula is valid for any non-self-intersecting (simple) polygon, which can be convex or concave". Using the general formula, the absolute value of the signed area is the area you wanted.
Let the point coordinates be
$ P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4). $
The area of triangle $ P_1P_2P_4 $ is
$ A := x_1(y_2-y_4) + x_2(y_4-y_1) + x_4(y_1-y_2). $ The area of triangle
$ P_4P_2P_3 $ is
$ B := x_4(y_2-y_3) + x_2(y_3-y_4) + x_3(y_4-y_2). $ Add the two signed areas to get $ A + B = (x_1y_2-x_2y_1) + (x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) + (x_4y_1-x_1y_4) $ from shoelace.
answered Jul 23 at 17:54
Somos
11.4k1833
add a comment |Â
up vote
0
down vote
accepted
If you use the formula given as sum of products, then this is a signed area. That is, depending on the order of the three points, you may get a negative value. The formula generalizes to any polygon. This is sometimes known as the shoelace formula. As the Wikipedia article states "The area formula is valid for any non-self-intersecting (simple) polygon, which can be convex or concave". Using the general formula, the absolute value of the signed area is the area you wanted.
Let the point coordinates be
$ P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4). $
The area of triangle $ P_1P_2P_4 $ is
$ A := x_1(y_2-y_4) + x_2(y_4-y_1) + x_4(y_1-y_2). $ The area of triangle
$ P_4P_2P_3 $ is
$ B := x_4(y_2-y_3) + x_2(y_3-y_4) + x_3(y_4-y_2). $ Add the two signed areas to get $ A + B = (x_1y_2-x_2y_1) + (x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) + (x_4y_1-x_1y_4) $ from shoelace.
answered Jul 23 at 17:54
Somos
11.4k1833
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If you use the formula given as sum of products, then this is a signed area. That is, depending on the order of the three points, you may get a negative value. The formula generalizes to any polygon. This is sometimes known as the shoelace formula. As the Wikipedia article states "The area formula is valid for any non-self-intersecting (simple) polygon, which can be convex or concave". Using the general formula, the absolute value of the signed area is the area you wanted.
Let the point coordinates be
$ P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4). $
The area of triangle $ P_1P_2P_4 $ is
$ A := x_1(y_2-y_4) + x_2(y_4-y_1) + x_4(y_1-y_2). $ The area of triangle
$ P_4P_2P_3 $ is
$ B := x_4(y_2-y_3) + x_2(y_3-y_4) + x_3(y_4-y_2). $ Add the two signed areas to get $ A + B = (x_1y_2-x_2y_1) + (x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) + (x_4y_1-x_1y_4) $ from shoelace.
answered Jul 23 at 17:54
Somos
11.4k1833
If you use the formula given as sum of products, then this is a signed area. That is, depending on the order of the three points, you may get a negative value. The formula generalizes to any polygon. This is sometimes known as the shoelace formula. As the Wikipedia article states "The area formula is valid for any non-self-intersecting (simple) polygon, which can be convex or concave". Using the general formula, the absolute value of the signed area is the area you wanted.
Let the point coordinates be
$ P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3), P_4=(x_4,y_4). $
The area of triangle $ P_1P_2P_4 $ is
$ A := x_1(y_2-y_4) + x_2(y_4-y_1) + x_4(y_1-y_2). $ The area of triangle
$ P_4P_2P_3 $ is
$ B := x_4(y_2-y_3) + x_2(y_3-y_4) + x_3(y_4-y_2). $ Add the two signed areas to get $ A + B = (x_1y_2-x_2y_1) + (x_2y_3-x_3y_2) +(x_3y_4-x_4y_3) + (x_4y_1-x_1y_4) $ from shoelace.
answered Jul 23 at 17:54
Somos
11.4k1833
edited Jul 24 at 0:54
answered Jul 23 at 17:54
Somos
11.4k1833
answered Jul 23 at 17:54
Somos
11.4k1833
answered Jul 23 at 17:54
Somos
11.4k1833
11.4k1833
add a comment |Â
add a comment |Â
up vote
0
down vote
You can surely use that method but it would be quite "bashy". Especially if you do it by hand. A better tool for this case (or any polygon) would be Gauss's Area Theorem, AKA shoelace theorem. I think this page provides a better explanation if shoelace theorem than Wikipedia.
https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem
answered Jul 23 at 18:01
Shrey Joshi
1389
add a comment |Â
up vote
0
down vote
You can surely use that method but it would be quite "bashy". Especially if you do it by hand. A better tool for this case (or any polygon) would be Gauss's Area Theorem, AKA shoelace theorem. I think this page provides a better explanation if shoelace theorem than Wikipedia.
https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem
answered Jul 23 at 18:01
Shrey Joshi
1389
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can surely use that method but it would be quite "bashy". Especially if you do it by hand. A better tool for this case (or any polygon) would be Gauss's Area Theorem, AKA shoelace theorem. I think this page provides a better explanation if shoelace theorem than Wikipedia.
https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem
answered Jul 23 at 18:01
Shrey Joshi
1389
You can surely use that method but it would be quite "bashy". Especially if you do it by hand. A better tool for this case (or any polygon) would be Gauss's Area Theorem, AKA shoelace theorem. I think this page provides a better explanation if shoelace theorem than Wikipedia.
https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem
answered Jul 23 at 18:01
Shrey Joshi
1389
answered Jul 23 at 18:01
Shrey Joshi
1389
answered Jul 23 at 18:01
Shrey Joshi
1389
answered Jul 23 at 18:01
Shrey Joshi
1389
1389
add a comment |Â
add a comment |Â
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the formula works for any triangle.Any quadrilateral can be split into two triangles but you have to be careful how you chose the diagonal. We have two diagonals, each creates two triangles.Calculate areas of four triangles and calculate area of a quadrilateral. You may get two different areas if quadrilateral is concave, the lesser one will give you the correct area. For convex quadrilateral, both areas will be the same.
– Vasya
Jul 23 at 17:21
Let's say the vertices are $V_1, V_2, V_3, V_4$, you need to calculate $A_triangleV_1V_3V_2$+$A_triangleV_1V_3V_4$ and $A_triangleV_2V_4V_1$+$A_triangleV_2V_4V_3$ then pick the minimum out of the two values.
– Vasya
Jul 23 at 17:29
Yes, you are right. In case of convex quadrilateral both these values will be same. They will be different for concave triangles and I can choose the lower value as the area.
– iKnowNothing
Jul 23 at 17:32